The current through an (ideal) inductor is continuous, just as the electric potential difference over a capacitor. So if the current \$ i_L \$ at \$t=0-\$ is zero, then so is the current at \$t=0+\$, immediately after the voltage source stepped to \$+5 V\$.
If you want a full solution of capacitor voltage and inductor current, you'd have to solve the differential equation for e.g. the capacitor voltage \$ u_C\$.
$$ u_S(t) = 5 * UnitStep $$
$$ i_R(t) = (u_S(t) - u_C(t))/R$$
$$ i_L(t) = \frac{1}{L}\int_0^t u_C(\tau) d\tau $$
$$ i_C(t) = C \frac{du_C}{dt} $$
In which the subscripts are quite self-explaining, except \$u_S(t)\$, which is the source voltage.
Now, omitting the time where possible, we can write $$u_c = u_S - R(i_L + i_C) = u_S - R\left(\frac{1}{L}\int u_C dt +C\frac{du_C}{dt}\right) $$
After differentiating the whole expression we get, and we write u for u_C,
$$ u'= -\frac{R}{L}u-RCu'' $$ or $$ RCu''+u'+\frac{R}{L}u = 0 $$
Substituting the values for \$R\$, \$L\$ and \$C\$ gives
$$ 150*10^{-8}u''+u'+ 150*10^3u = 0 $$
Multiplying with \$2*10^6\$ gives us
$$ 3u''(t) + 2 * 10^6 u'(t) + 3 * 10^{11} u(t) = 0 $$
Some boundary conditions of this problem are \$u(0)\$ and \$u'(0)\$. Of course \$u(0)=0\$ as at the start the capacitor is empty. \$u'(0)\$ can be found by realizing ourselves that \$i_L(0) = 0\$ because the current in any self inductance is continuous and therefore all the current at \$t=0\$ will flow into the capacitor. This current is caused by the voltage drop of \$5 V\$ over the resistors when the step is enacted and C has zero charge hence zero voltage. So this current is \$ i_C(0+) = 5/150 A \$ which causes a rate of change of the capacitor voltage \$ u'_C(0+) = C*i_C(0+) = \frac{5}{150}*10^{-8} V/s\$. Therefore the boundary conditions are
$$ u(0) = 0 $$ $$ u'(0) = \frac{1}{3} * 10^7 $$
The exact solution of this differential equation with boundary conditions is
$$ u_C(t) = 5 * \sqrt 10 * e^{-10^5 (10 + \sqrt 10) t/3} (e^{2*10^5 \sqrt(10) t/3} - 1) $$
and it plots like this.
Now this deals with the voltage \$u_C(t)\$ over the capacitor and your question was regarding the inductor current which is $$ i_L(t) = \frac{1}{L} \int_0^t u(\tau) d\tau $$ and with \$a=\sqrt 10\$, \$b=100000/3\$ (and \$a^2=10\$) this becomes
$$i_L(t) = \frac{1 - e^{-a^2 b t} \left(cosh(a b t) + a sinh(a b t)\right)}{b(1 - 1/a^2)} $$
Of course, when \$t=0\$, the exponent becomes \$e^{-0}=1\$, \$sinh(0)=0\$ and \$cosh(0)=1\$, hence the whole expression becomes $$ i_L(0+) = \frac{1-1}{b(1-1/a^2)}=0 $$
